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not right :(

  t^5 + 1/t^5 = (1 + 1/t)(t^4 - t^3 * 1/t + t^2 * 1/t^2 - t * 1/t^3 + 1/t^4)

= (t + 1/t)(t^4 + 1/t^4 -(t^2 +1/t^2) + 1)

Lets stop there,

Now we need to find t^4 + 1/t^4 and t^2 + 1/t^2

t^2 + 1/t^2 = (t + 1/t)^2 - t * 1/t

t^2 + 1/t^2 = 3^2 - 1

= 9 - 1 = 8

t^4 + 1/t^4:

(t^2 + 1/t^2)^2 = t^4 + 2 + 1/t^4

Soo, t^4 + 1/t^4 = 8^2 - 2

t^4 + 1/t^4 = 64 - 2 = 62

Now we can plug this back!

(t + 1/t)(62 - 8 + 1) = 3 * 55 = 165

I love these types of questions and I kinda just reviewed this yesterday!

I hope you understand!

The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.

x + y + 3z  =  10     --->     x -2     --->     -2x - 2y - 6z  =  -20

-4x + 3y + 5z  =  7       --->                      -4x + 3y + 5z  =  7

Add down the columns:                             -6x - z  =  -13   

Since  kx + z = 3        --->         z  =  3 - kx

Substituting these:        -6x - (3 - kx)  =  -15

                                         (k - 6)x  = -18

If  k - 6  =  0,  there is no value for x that will result in a product of -18

so  k = 6  results in no possible solution.

For every other value of k, there will be a value of x that can produce a product of -18.

So the answer is k = 6.

x = 5 pt questions

30 -x   =  2 point questions

5x   +    2 ( 30 - x) = 114

3x = 54

x = 18     5-pointers     then  12     2-pointers        

No I don't. Please Explain. Thank you!

We can set up we have 2 variables for this problem. 

x = # of 5 point questions

y = # of 2 point questions

Since we have 2 variables, we need 2 equations, and we have those equations.

We know that...

\(x + y = 30\)

\(5x + 2y = 114\)

To solve, we first have to make it a one variable equation.

Since \(x = 30 - y\), we can replace all of the \(y\) with \(30 - x\).

So \(5x + 2y = 114\) changes to \(5x + 2(30-x) = 114\)

\(5x + 2(30-x) = 114\)

\(5x + 60 - 2x = 114\)

\(3x + 60 = 114\)

\(3x = 54\)

\(x = 18\)

So...

\(y = 12\)

Your answer is...

There are 18 5 point questions and 12 2 point questions.

And you are done! :)

Easy, if you know the formula. 

Hint:

x^5 + y^5 = ?

If you still don't know, please reply!

I'll help you get started~

Set up an equation~

11x + 5 = 5x + 59

        6x = 54

           x=9

11x + 5 = 99 + 5 = 104

     z = 180 - 104 = 76

so...

x = 9°

z = 76°

~

18/14 = x/12

9/7 = x/12

x = 108/7

Ohh, nice job. :))

=^._.^=

Nevermind, I got it right, I just forgot about the 2...

$\left( \frac{4}{x} \right)^{-1} \left( \frac{3x^3}{x} \right)^2 \left( \frac{1}{2x} \right)^{-3} = \frac{x}{4} \cdot (3x^2)^2 \cdot (2x)^3 = \frac{x}{4} \cdot 9x^4 \cdot 8x^3 = \boxed{18x^8}$. 

Hint~

12 + 3 = 15

We can set up a proportion~

x / 35 = 12 / 15

Cross multiply~

15x = 420

______________

x = 28

~

When ab is divided by n, the remainder is 2.

The answer isn't right, but thanks for trying

~

You can use the formula

\[\sum_{k = 1}^n k = \frac{n(n + 1)}{2}\]

\[\sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}\]

Then $a_8 + a_9 + \dots + a_{23} = \boxed{326520}$.

That is correct, thanks 

sin ( x ) = 2.4 / 4.8

= 1 / 2

sin ( 30 ) = 1 / 2

So...

 x = 30

~

i = v/r

calculate r

200 = 50/r

r = 50/200  

cutting wire in half makes new  r = 1/2 r

i = 300/(1/2(50/200)) = 2400 mAmps     or   2.4 amps

Im sorry but thats incorrect

You will measure 200*300/50 = 1200 milliamps.

For a forty by forty grid of squares, the total length of all line segments is 3460.

Smallest: 13*8 = 104

Largest: 13*76 = 988

76 - 7 =69

True, True, 

I assume DP is decimal point

And if you round to 1 decimal point, it would be about...

282.7

(it would be 282.6 if you use 3.14 though)

~

You sure?